II Institute of , LMU Munich – Winter Term 2011/2012 Prof. T. Ø. Sørensen Ph.D, A. Michelangeli Ph.D

HOMEWORK ASSIGNMENT no. 9, issued on Wednesday 14 December 2011 Due: Wednesday 21 December 2011 by 2 pm in the designated “FA2” box on the 1st floor Info: www.math.lmu.de/~michel/WS11-12_FA2.html

Each exercise sheet is worth a full mark of 40 points. Correct an- swers without proofs are not accepted. Each step should be justified. You can hand in the solutions either in German or in English.

Exercise 33. (Spectrum of a partial isometry.) (i) Let H be a and let T ∈ B(H) with kT k 6 1. Show that O 6 TT ∗ 6 1 (in the sense of operators) and that the√ operator M : H⊕H → H⊕H, x⊕y 7→ (T x+Sy)⊕0 is a partial isometry, where S := 1 − TT ∗.(Hint: Problem 24.) (Note: for you own curiosity, identify the initial and final spaces of the partial isometry M [you should be able to do that] and see that they are quite nasty. In fact, the definition of partial isometries is deceptively simple and the standard examples [orthogonal projections, unitary operators, etc.] continue the deception, but the structure of a partial isometry can be quite complicated!) (ii) Let K be a compact subset of the closed disc {λ ∈ C | |λ| 6 1} such that 0 ∈ K. Show that there exists a partial isometry U on a Hilbert space H with spectrum K. (Hint: part (i) and Exercise 18 (iv).)

Exercise 34. (Examples of polar decompositions.) (i) Let H be a Hilbert space, η, ξ ∈ H with kηk = kξk = 1, and T = |ξihη| : H → H (i.e., T is the operator defined by T x := hη, xiξ for all x ∈ H). Find the polar decomposition T = UT |T | of T , i.e., give the partial isometry UT and the operator |T |. 2 2 (ii) Find the polar decomposition R = UR|R| of the right R : ` (N) → ` (N), 2 R(x1, x2, x3,... ) := (0, x1, x2,... ) for all x = (x1, x2, x3,... ) ∈ ` (N). 2 2 (iii) Find the polar decomposition L = UL|L| of the left shift operator L : ` (N) → ` (N), 2 L(x1, x2, x3,... ) := (x2, x3, x4 ... ) for all x = (x1, x2, x3,... ) ∈ ` (N). | | 2 → (iv) Find the polar decompositionR V = UV V of the Volterra integral operator V : L [0, 1] 2 x ∈ L [0, 1], (Vf)(x) := 0 f(y)dy for almost all x [0, 1]. Important: the full solution here is to produce the explicit “closed” form of the operators UV and |V |. Giving the canonical decomposition of |V | is not enough, show that in fact |V | is an integral operator and UV is a (with an explicit, “easy” form).

1 Exercise 35. (The discrete Laplacian.) 2 2 2 Consider the operator ∆ : ` (Z) → ` (Z) defined on every x = (. . . , xn−1, xn, xn+1,... ) ∈ ` by X (∆x)n := (xn − xm) (n ∈ Z) . m∈Z |m−n|=1 (i) Show that ∆ is bounded and self-adjoint. (ii) Show that O 6 ∆ 6 4 · 1. (iii) Compute k∆k. (iv) Determine σ(∆). (Hint: Problem 12 (i), Exercise 18 (iii).) (v) Produce a measure space (M, µ), an isomorphism U : `2(Z) → L2(M, dµ), and a function F :M → R such that U∆ U −1 acts on L2(M, dµ) as the operator of multiplication by F .

Exercise 36. (The spectrum is upper semicontinuous, but not continuous.) (i) Let X be a and let T ∈ B(X). Show that for every bounded open set Ω ⊂ C such that σ(T ) ⊂ Ω there exists ε > 0 such that if S ∈ B(X) with kS − T k 6 ε then σ(S) ⊂ Ω. (ii) Consider the operators A(N),A ∈ B(`2(Z)), N ∈ N, defined on the canonical orthonormal 2 basis {en}n∈Z of ` (Z) by ( ( 6 6 (N) en+1 if n = 0 en+1 if n = 0 A en := 1 , A en := , N e1 if n = 0 0 if n = 0 and then extended by linearity and boundedness. Show that k k • A(N) −−−→ A, N→∞ • σ(A(N)) ⊂ {λ ∈ C | |λ| = 1} for all N ∈ N, • σ(A) contains points λ with |λ|= 6 1, so that the spectrum of A is discontinuously different from the spectra of the A(N)’s. Optional: show that in fact σ(A(N)) is the unit circle and σ(A) is the unit disk of C.

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